Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $f(x) = o(\sqrt{x})$ as $x\rightarrow\infty$ and let $x^*(a)$ denote the minimizer of $f(x) + a^{3/2}/x$, that is, the value of $x$ that minimizes said expression (assuming such a value exists). As $a\rightarrow\infty$, is it true that $x^*(a) = \omega(a)$, i.e. that the minimizer grows super-linearly in $a$?

share|improve this question
    
What is a "minimizer" in this context? –  Henning Makholm Jul 15 '12 at 1:29
    
The value of $x$ that minimizes $f(x) + a^{3/2}/x$. Of course, the expression may not have a minimizer (e.g. if f(x) = 0) –  Charlie Jul 15 '12 at 1:30

1 Answer 1

up vote 0 down vote accepted

Yes. It suffices to show that the minimum is $o(\sqrt{a})$. Suppose not. Then there is $c>0$ and a sequence $a_n\to\infty$ such that $f(x)+a_n^{3/2}/x\ge c\sqrt{a_n}$ for all $n$ and all $x$. But for large $x$ we have $f(x)<c^{3/2}\sqrt{x}/1000$. Setting $x=100a_n/c$ yields a contradiction.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.